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Question:

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by:

E0r(r+r1).lL

LE0rlr1

LE0r(r+r1)l

E0lL

Solution:

The current in the circuit ,I=E0/(r+r1)
The potential drop across the wire is V=Ir=E0r/(r+r1)
Potential gradient,k=V/L=E0r/(r+r1)L
Thus, emf E=kl=E0rl/(r+r1)L